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Use the Corresponding Angles Theorem and the Angle-Angle (AA) Similarity Postulate. Then, rewrite the obtained proportion by using the Segment Addition Postulate.
See solution.
Let's consider △ ACD and BE in a manner that BE ∥ CD and BE intersects the other two sides of the triangle.
Recall the definition of the Corresponding Angles Theorem.
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Corresponding Angles Theorem |
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If parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. |
As we can see, â–³ ACD and â–³ ABE have two pairs of congruent angles. Now, recall the Angle-Angle (AA) Similarity Postulate.
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Angle-Angle (AA) Similarity Postulate |
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If two angles of a triangle are congruent to two angles of another triangle, then the triangles are similar. |
Thanks to this, we know that â–³ ACD ~ â–³ ABE. Because of that, their corresponding side lengths are proportional. AC/AB = AD/AE Let's rewrite the numerators by using the Segment Addition Postulate. AC = AB + BC AD = AE + ED Next, let's substitute the equations above into the proportion equation.
AC= AB+BC, AD= AE+ED
Write as a sum of fractions
Calculate quotient
LHS-1=RHS-1
LHS * AB* AE=RHS* AB* AE
.LHS /(BC* ED).=.RHS /(BC* ED).
Rearrange equation
We have obtained the required proportion.
Given: & BE∥ CD
Prove: & ABBC = AEED
Proof: By the Corresponding Angles Theorem we get that ∠ABE ≅ ∠ACD and ∠AEB ≅ ∠ADC. Thus, △ ACD~△ ABE thanks to the AA Similarity Postulate. Then, we get the proportion ACAB = ADAE. The Segment Addition Postulates gives us AC = AB + BC and AD = AE + ED.
Substituting the last two equations into the proportion and performing the required operations, we get that ABBC = AEED, which is what we wanted.